User:EML4500.f08.RAMROD.E/HW3

Lecture Notes: Meeting 14 - 9/26/08
In order to relate the axial coordinate system's forces, f(e), to the element coordinate system's axial forces, p(e), it is necessary to employ a transformation matrix, T(e) such that:

$$\begin{Bmatrix}

P^{(e)}_{1}\\ P^{(e)}_{2} \end{Bmatrix} = T^{(e)}*

\begin{Bmatrix}

f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

Looking at this equation, we can tell that T(e) must be a 2X4 matrix in order for the dimensions to work out.

Knowing this, and using the element F-D relationship,

$$\hat{k}^{(e)}*q^{(e)} = p^{(e)}$$

we can substitute in the Transformation Matrix relationship above to get an equation relating the cartesion forces and displacements via the transformation matrix and \hat{k}. The resulting equation is shown below.

$$\hat{k}^{(e)}*(T^{(e)}*d^{(e)}) = T^{(e)}$$

Rearranging this equation requires moving the T(e) term to the left side to allow us to solve for the internal forces in f(e). Since T(e) is rectangular, the transpose, rather than the inverse, is multiplied by each side resulting in:

$$[T^{(e)T}*\hat{k}^{(e)}*(T^{(e)}]*d^{(e)} = f^{(e)}$$

In order for this to be true, it must be proven that,

$$k^{(e)}= [T^{(e)T}*\hat{k}^{(e)}*(T^{(e)}]$$

This proof is given below as the assigned homework.

Now that we have proven the derivation of k^e, we can continue to solving for the global displacement degrees of freedom.

Using the principle of virtual work (i.e. d1 = d2 = d5 = d6 = 0) we can reduce the global equation to two unknown DOF's, d3 and d4.

The next homework assignment was to solve for the eigenvalues of the global stiffness matrix, K.